Problem: $h(n) = -6n^{2}$ $f(n) = -3n^{2}+2n+3+2(h(n))$ $ h(f(0)) = {?} $
Solution: First, let's solve for the value of the inner function, $f(0)$ . Then we'll know what to plug into the outer function. $f(0) = -3(0^{2})+(2)(0)+3+2(h(0))$ To solve for the value of $f$ , we need to solve for the value of $h(0)$ $h(0) = -6(0^{2})$ $h(0) = 0$ That means $f(0) = -3(0^{2})+(2)(0)+3+(2)(0)$ $f(0) = 3$ Now we know that $f(0) = 3$ . Let's solve for $h(f(0))$ , which is $h(3)$ $h(3) = -6(3^{2})$ $h(3) = -54$